Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
At the moment the switch is closed, the magnetic field increases while half of the loop is inside this field.
Let's assume that the loop was initially at rest, then the increase in the magnetic field creates an induced emf that creates a current which will produce a magnetic field that opposes this increase.
And since the original magnetic field is out of the page, then the induced flux must be into the page.
This means that the current in the loop, according to the right-hand rule, must be clockwise.
Hence, at least 3 wires of the loop experience a magnetic force. The force on the front wire is to the left while the forces on the upper and lower wire are canceling each other. The fourth wire experiences no force since it is not exposed to the magnetic flux.
We know that the induced current is given by
$$I_{\rm loop}=\dfrac{\varepsilon_{\rm loop }}{R_{\rm loop}}\tag 1$$
We know that the induced emf $\varepsilon$ is given by
$$\varepsilon_{\rm loop}=\left|\dfrac{d\Phi_{\rm }}{dt}\right|\tag 2$$
where $ \Phi =\vec A\cdot \vec B =AB\cos90^\circ=BA $ where $A$ is constant (just at the moment the switch is closed) while $B$ changes.
Noting that the force exerted on the loop will kick it out of the field but at the moment we turn on the switch it may move a negligible distance that we can ignore it.
Plug into (2),
$$\varepsilon_{\rm loop}=A \left|\dfrac{d B}{dt} \right| $$
where $A=\frac{1}{2}l^2$, where $l$ is the length of one side of the loop.
$$\varepsilon_{\rm loop}=\frac{1}{2}l^2\left|\dfrac{\Delta B}{\Delta t} \right| $$
$$\varepsilon_{\rm loop}=\frac{1}{2}l^2\left|\dfrac{ B_f-B_i}{\Delta t} \right| $$
Plug into (1),
$$I_{\rm loop}=\dfrac{l^2}{2R_{\rm loop}}\left|\dfrac{ B_f-B_i}{\Delta t} \right| $$
Plug the known;
$$I_{\rm loop}=\dfrac{(0.08)^2}{2(0.01)}\left|\dfrac{ 1-0}{(0.01)} \right| $$
$$I_{\rm loop}=\color{red}{\bf 32}\;\rm A$$
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$$\color{blue}{\bf [b]}$$
Recalling that the magnetic force exerted on a current-carrying wire is given by
$$F_B=BIl$$
And since we assumed that the magnetic field changes uniformly, $B_{\rm avg}=\dfrac{\Delta B}{2}=\dfrac{1-0}{2}=0.5\;\rm T$
Hence, the average force exerted on the leading edge (front wire) is given by
$$F_{B,\rm avg}=B_{\rm avg}Il$$
As the author mentioned in the hint, we need to use the impulse to find the speed of the loop after being kicked out.
$$F_{B,\rm avg}\Delta t=\Delta p=mv_f=mv_i$$
So,
$$ B_{\rm avg}Il\Delta t=mv_f-0$$
Thus,
$$v_f=\dfrac{ B_{\rm avg}Il\Delta t}{m}$$
Plug the known;
$$v_f=\dfrac{ (0.5)(32)(0.08)(0.01)}{(0.01)}$$
$$v_f=\color{red}{\bf 1.28}\;\rm m/s$$
